\section{Q2: A reflection theorem  + A Barrier Option}
Let
\begin{align}
p = 1 - \frac{2r}{\sigma^2}
\end{align}
\subsection{Q.2a}
We let $Z(t) = \paren{S(t)/H}^p$ and show that $Z(t)/Z(0)$ is a positive, mean-1 $\mQ$-martingale. We accomplish this by calculating the $\mQ$-dynamics of $Z(t)$.\newline

By It\=o we have
\begin{align}
dZ(t) & = \frac{p}{H} \left( \frac{S(t)}{H}\right)^{p-1}dS + \half \frac{p(p-1)}{H^2} \left( \frac{S(t)}{H} \right)^{p-2}dS^2\nonumber\\
& = \frac{p}{H} \left( \frac{S(t)}{H}\right)^{p-1} \left( rS(t) dt + \sigma S(t) \dw\right) + \half \frac{p(p-1)}{H^2} \left( \frac{S(t)}{H} \right)^{p-2}\sigma^2 S(t)^2dt\nonumber\\
& = p\left(\frac{S(t)}{H}\right)^p\left( r + \half (p-1)\sigma^2\right)dt + p\sigma \left(\frac{S(t)}{H}\right)^p \dw\nonumber\\
& = p\sigma \left(\frac{S(t)}{H}\right)^p \dw = p\sigma Z(t) \dw. \label{eq:girsanov}
\end{align}
By Lemma 4.9 of Bj\"ork $Z(t)/Z(0)$ is a $\mQ$-martingale. It must be positive since $S(t)$ is. Using the tower property of conditional expectations we have

\begin{align}
E\brackk{\frac{Z(t)}{Z(0)}} & = E\brackk{E\brackk{\frac{Z(t)}{Z(0)}\lvert\mathcal{F}_0}}=\frac{1}{Z(0)}E\brackk{E\brackk{Z(t)\lvert\mathcal{F}_0}} = E\brackk{Z(0)}= \frac{Z(0)}{Z(0)} = 1.\nonumber
\end{align}
Thus we also have
\begin{align}
d\mQ^Z = \frac{Z(T)}{Z(0)}d\mQ =L(T)\mQ,
\end{align}
which follows from Girsanovs Theorem (Bj\"ork theorem 11.3).
\subsection{Q.2b}
It follows from the previous question that
\begin{align}
\pi^{\hat{g}} (t) & = e^{-r(T-t)}E^{\mQ}_t\brackk{\paren{\frac{S(T)}{H}}^p g\paren{\frac{H^2}{S(T)}}}\nonumber\\
& = e^{-r(T-t)}E^{\mQ}_t\brackk{\paren{L(T) g\paren{\frac{H^2}{S(T)}}}}\nonumber\\
& = e^{-r(T-t)}E^{\mQ}_t\brackk{\paren{\frac{S(T)}{H}}^p}  E^{\mQ^Z}_t\brackk{g\paren{\frac{H^2}{S(T)}}}\label{eq:bayes}\\
& = e^{-r(T-t)}\paren{\frac{S(t)}{H}}^p  E^{\mQ^Z}_t\brackk{g\paren{\frac{H^2}{S(T)}}}\label{eq:z_mq}
\end{align}
where in~\eqref{eq:bayes} we made use Bj\"ork proposition B.41 and and in~\eqref{eq:z_mq} that $Z(t)$ is a $\mQ$-martingale.
\subsection{Q.2c}
Let $Y(t) = H^2/S(t)$. We show that
\begin{align}
dY(t) = rY(t)dt - \sigma Y(t)\dw^{\mQ^Z}.
\end{align}
First note that $\dw^{\mQ^Z} = \dw^{\mQ} - p\sigma dt$, see Bj\"ork theorem 11.3 (Girsanovs Theorem).
The result follows. Indeed, from It\=o we have the following (a little short this time, since we just did one)
\begin{align}
dY(t) & = \frac{H^2}{S(t)}\paren{\paren{\sigma^2 - r}dt - \sigma \dwq}\nonumber\\
& = \frac{H^2}{S(t)}\paren{\paren{\sigma^2 - r}dt - \sigma \paren{ p\sigma dt + \dwqz}}\label{q2c:girsanov}\\
& = rY(t)dt - \sigma Y(t) \dwqz,
\end{align}
where in~\eqref{q2c:girsanov} we made use of the Girsanov kernel, $p\sigma$, of the measure transformation. Thus the distribution of $Y$ under $\mQ^Z$ is the same as $S$ under $\mQ$. Hence anything and everything we say about $Y$ with respect to its distribution under $\mQ^Z$ must the same for $S$ under $\mQ$. In particular
\begin{align}
\pi^{\hat{g}}(t) &= e^{-r(T-t)}\paren{\frac{S(t)}{H}}^p E^{\mQ^Z}_t\brackk{g\paren{Y(T)}}\nonumber\\
& = e^{-r(T-t)}\paren{\frac{S(t)}{H}}^p E^{\mQ}_t\brackk{g\paren{H^2/S(t)}} = e^{-r(T-t)}\paren{\frac{S(t)}{H}}^p f\paren{\frac{H^2}{S(T)}}
\end{align}

\subsection{Q.2d}
Consider a strike-K call-option, with the additional provision that if the underlying reaches a barrier B, which is less than K, the option becomes worthless.
Suppose we buy the standard call and sell the reflected-through-B claim. If the barrier is hit we liquidate the portfolio. Hence the pay-off from the contingent claims (of which we hold none), at time $T$, is zero. Furthermore note that the portfolio also has value zero if the barrier is hit. Indeed, given that $S(t) = B$ the price is given by
\begin{align}
\pi^{g}(t) - \pi^{\hat{g}}(t) & = e^{-r(T-t)}\paren{f\paren{S(t)}-\paren{\frac{S(t)}{B}}^p f\paren{\frac{B^2}{S(t)}}}\nonumber\\
& = e^{-r(T-t)}\paren{f\paren{B}-\paren{1}^p f\paren{B}} = 0.
\end{align}
This covers the case where the barrier is hit.
If the barrier is not hit we have the following pay-off at time $T$
\begin{align}
\paren{S(T)-K}^+ - \paren{\frac{S(T)}{B}}^p\paren{\frac{B^2}{S(T)}-K}^+ = \paren{S(T)-K}^+.
\end{align}
Indeed, this follows since $S(T) > B$ and thus we have $B^2/S(T) < B < K$.
The initial price is given by
\begin{align}
\pi^{g}(0) - \pi^{\hat{g}}(0) = &s\Phi(d_1(0,S(0))) - e^{-r(T-t)}K\Phi(d_2(0,S(0)))\nonumber\\
 & - \paren{\frac{S(0)}{B}}^p\paren{\frac{B^2}{S(0)}\Phi(d_1(0,B^2/S(0))) + e^{-r(T-t)}K\Phi(d_2(0,B^2/S(0)))},
\end{align}
where 
\begin{align}
d_{1/2}(t,s) & = \frac{\log (s/K) + \paren{r \pm \half\sigma^2}\paren{T-t}}{\sqrt{T-t}\sigma}.\nonumber
\end{align}
If $r=0$ we have the following pay-off the reflected through $B$-claim
\begin{align}
\paren{\frac{S(T)}{B}}^p\paren{\frac{B^2}{S(T)}-K}^+ & =\paren{\frac{S(T)}{B}}\paren{\frac{B^2}{S(T)}-K}^+\nonumber\\
& =\paren{B-K\frac{S(T)}{B}}^+\nonumber\\
& =\frac{K}{B}\paren{\frac{B^2}{K}-S(T)}^+.\nonumber\\
\end{align}
Thus the claim is simply $\frac{K}{B}$ put options on $S$ with strike $\frac{B^2}{K}$ and expiry $T$.
\newpage
